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prove that: sinA-cosA+1/sinA+cosA-1=1/secA-tanA  
Asked by banerjee_milky | 16 Feb, 2019, 11:23: AM
answered-by-expert Expert Answer
For conveneience let us use the substitution, sinA = s and cosA = c
 
begin mathsize 12px style L H S space equals space fraction numerator sin A space minus cos A plus 1 over denominator sin A plus cos A minus 1 end fraction space equals space fraction numerator s minus open parentheses c minus 1 close parentheses over denominator s plus left parenthesis c minus 1 right parenthesis end fraction space L H S space equals fraction numerator begin display style s minus open parentheses c minus 1 close parentheses end style over denominator begin display style s plus left parenthesis c minus 1 right parenthesis end style end fraction cross times fraction numerator s minus open parentheses c minus 1 close parentheses over denominator s minus open parentheses c minus 1 close parentheses end fraction space equals space fraction numerator begin display style open square brackets s minus open parentheses c minus 1 close parentheses close square brackets squared end style over denominator s squared minus left parenthesis c minus 1 right parenthesis squared end fraction space equals space fraction numerator s squared plus c squared minus 2 c plus 1 minus 2 s c plus 2 s over denominator s squared minus c squared minus 1 plus 2 c end fraction space l e t space u s space u s e space t h e space i d e n t i t y space s squared plus c squared equals 1 space a n d space s squared minus 1 space equals space minus c squared space space i n space t h e space a b o v e space e x p r e s s i o n L H S space equals space fraction numerator 2 minus 2 c plus 2 s minus 2 s c over denominator 2 c minus 2 c squared end fraction space equals space fraction numerator open parentheses 1 minus c close parentheses begin display style open parentheses 1 plus s close parentheses end style over denominator c open parentheses 1 minus c close parentheses end fraction space equals space fraction numerator 1 plus s over denominator c end fraction space equals space s e c A plus tan A end style
begin mathsize 12px style R H S space equals space fraction numerator 1 over denominator s e c A space minus space tan A end fraction space equals space space fraction numerator begin display style 1 end style over denominator begin display style s e c A space minus space tan A end style end fraction cross times fraction numerator s e c A plus tan A over denominator s e c A plus tan A end fraction space equals space fraction numerator s e c A space plus space tan A over denominator s e c squared A space minus space tan squared A end fraction space equals space s e c A space plus space tan A space open square brackets because space s e c squared A space minus space tan squared A space equals space 1 space close square brackets end style
hence we have proved LHS = RHS,
Answered by Thiyagarajan K | 16 Feb, 2019, 12:44: PM

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